Consider a circular conductor with radius a that carries a current I. Find the magnetic field at point P on the axis of the loop, at a distance x from the center.

From Biot-Savart Law:

$$\vec{B} = \frac{\mu_{0}}{4\pi} \int \frac{I}{r^{2}} \, d\vec{l} \times \hat{r}$$

Due to the symmetry of the current loop, the $\vec{B_{y}}$ at point P due to the top half of the current loop will cancel out the $\vec{B_{y}}$ due to the lower half of the current loop. Hence, we will only need $\vec{B_{x}}$.

$$dB_{x} = \frac{\mu_{0}}{4\pi} \frac{I}{x^{2}+a^{2}} \, dl \, \cos{\theta}$$

From the diagram,

$$\begin{aligned} \cos{\theta} &= \frac{a}{r} \\ &= \frac{a}{\sqrt{x^{a}+a^{a}}} \end{aligned}$$

Hence,

$$\begin{aligned} B_{x} &= \frac{\mu_{0}Ia}{4 \pi \left( x^{2} + a^{2} \right)^{\frac{3}{2}}} \int\limits_{0}^{2 \pi a} \, dl \\ &= \frac{\mu_{0}Ia^{2}}{2 \left( x^{2} + a^{2} \right)^{\frac{3}{2}}} \end{aligned}$$

If there are a coil consisting of N loops, all with the same radius,

$$B_{x} = \frac{\mu_{0}NIa^{2}}{2 \left( x^{2} + a^{2} \right)^{\frac{3}{2}}}$$

Note that the answer above assumes that the loops have no thickness.

From the equation above, it is obvious that the maximum $B_{x}$ will be at $x = 0$.

$$B_{max} = \frac{\mu_{0} N I}{2a}$$

Alternatively, you can set $\frac{dB_{x}}{dx} = 0$.

$$\frac{dB_{x}}{dx} = \, – \frac{3 \mu_{0} NIa^{2}x}{2 \left( x^{2} + a^{2} \right)^{\frac{5}{2}}} = 0$$

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